tex: finalize NFDBP demo

latex/content.tex

@@ -312,25 +312,25 @@ new constraints on the first bin can be expressed as follows :
 
 \subsection{La giga demo}
 
-Let $ k \in \mathbb{N} $. Let $ (U_n)_{n \in \mathbb{N}} $ be a sequence of
+Let $ k \geq 2 $. Let $ (U_n)_{n \in \mathbb{N}^*} $ be a sequence of
 independent random variables with uniform distribution on $ [0, 1] $, representing
 the size of the $ n $-th item.
 
 Let $ i \in \mathbb{N} $. $ T_i $ denotes the number of items in the $ i $-th
 bin. We have that
 
-\begin{equation}
+\begin{equation*}
 	T_i = k \iff U_1 + U_2 + \ldots + U_{k-1} < 1 \text{ and } U_1 + U_2 + \ldots + U_{k} \geq 1
-\end{equation}
+\end{equation*}
 
 Let $ A_k = \{ U_1 + U_2 + \ldots + U_{k} < 1 \}$. Hence,
 
-\begin{align*}
-	% TODO = k
+\begin{align}
+	\label{eq:prob}
 	P(T_i = k)
 	 & = P(A_{k-1} \cap A_k^c)                                           \\
 	 & = P(A_{k-1}) - P(A_k) \qquad \text{ (as $ A_k \subset A_{k-1} $)} \\
-\end{align*}
+\end{align}
 
 We will try to show that $ \forall k \geq 1 $, $ P(A_k) = \frac{1}{k!} $. To do
 so, we will use induction to prove the following proposition \eqref{eq:induction},
@@ -373,20 +373,29 @@ $ f_{U_{k-1}}(y) = 1 $. We can then integrate by parts :
 
 \begin{align*}
 	P(S_k < a)
-	 & = \int_{0}^{a} f_{S_{k-1}}(x) \cdot \left( \int_{0}^{a-x} 1 dy \right) dx   \\
-	 & = \int_{0}^{a} f_{S_{k-1}}(x) \cdot (a - x) dx                              \\
-	 & = a \int_{0}^{a} f_{S_{k-1}}(x) dx - \int_{0}^{a} x f_{S_{k-1}}(x) dx       \\
+	 & = \int_{0}^{a} f_{S_{k-1}}(x) \cdot \left( \int_{0}^{a-x} 1 dy \right) dx        \\
+	 & = \int_{0}^{a} f_{S_{k-1}}(x) \cdot (a - x) dx                                   \\
+	 & = a \int_{0}^{a} f_{S_{k-1}}(x) dx - \int_{0}^{a} x f_{S_{k-1}}(x) dx            \\
 	 & = a \int_0^a F'_{S_{k-1}}(x) dx - \left[ x F_{S_{k-1}}(x) \right]_0^a
-	+ \int_{0}^{a} x F_{S_{k-1}}(x) dx \qquad \text{(IPP)}                         \\
+	+ \int_{0}^{a} F_{S_{k-1}}(x) dx \qquad \text{(IPP : }x, F_{S_{k-1}} \in C^1([0,1]) \\
 	 & = a \left[ F_{S_{k-1}}(x) \right]_0^a - \left[ x F_{S_{k-1}}(x) \right]_0^a
-	+ \int_{0}^{a} \frac{x^{k-1}}{(k-1)!} dx                                       \\
-	 & = \left[ \frac{x^k}{k!} \right]_0^a                                         \\
-	 & = \frac{a^k}{k!}                                                            \\
+	+ \int_{0}^{a} \frac{x^{k-1}}{(k-1)!} dx                                            \\
+	 & = \left[ \frac{x^k}{k!} \right]_0^a                                              \\
+	 & = \frac{a^k}{k!}                                                                 \\
 \end{align*}
 
-We have shown that $ (\mathcal{H}_{k}) $ is true, so by induction, $ \forall k \geq 1 $,
+\paragraph{Conclusion} We have shown that $ (\mathcal{H}_{k}) $ is true, so by induction, $ \forall k \geq 1 $,
 $ \forall a \in [0, 1] $, $ P(U_1 + U_2 + \ldots + U_{k} < a) = \frac{a^k}{k!} $. Take
-$ a = 1 $ to get $ P(U_1 + U_2 + \ldots + U_{k} < 1) = \frac{1}{k!} $.
+$ a = 1 $ to get
+
+\[ P(U_1 + U_2 + \ldots + U_{k} < 1) = \frac{1}{k!} \]
+
+Finally, plugging this into \eqref{eq:prob} gives us
+
+\[
+	P(T_i = k) = P(A_{k-1}) - P(A_{k}) = \frac{1}{(k-1)!} - \frac{1}{k!} \qquad \forall k \geq 2
+\]
+